What is the integral of (tan(x))^3(sec(x))^3 ?

The integral of (tan(x))^3(sec(x))^3 is (sec^5(x))/5-(sec^3(x))/3+C

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integral of (tan(x))^3(sec(x))^3

\int (tan (x))^{3} (sec (x))^{3} d x

\frac{sec ^{5} ( x )}{5} - \frac{sec ^{3} ( x )}{3} + C

Solution steps

\int (tan (x))^{3} (sec (x))^{3} d x

Simplify

= \int tan^{3} (x) sec^{3} (x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (x)) tan (x) sec^{3} (x) d x

Apply u-substitution

: \int u^{2} (u^{2} - 1) d u

= \int u^{2} (u^{2} - 1) d u

Expand

u^{2} (u^{2} - 1) : u^{4} - u^{2}

= \int u^{4} - u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{4} d u - \int u^{2} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{2} d u = \frac{u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{u ^{3}}{3}

Substitute back

u = sec (x)

= \frac{sec ^{5} ( x )}{5} - \frac{sec ^{3} ( x )}{3}

Add a constant to the solution

= \frac{sec ^{5} ( x )}{5} - \frac{sec ^{3} ( x )}{3} + C

\int (u + 3 u - 2 u^{3}) d u

\int \frac{5 x}{3 x ^{2} - 5} d x

\int z 1 - z^{2} d z

\int - sin (x) sec^{3} (x) d x

\int \frac{x ^{3} + 3 x ^{2} - 9 x - 2}{x - 2} d x

What is the integral of (tan(x))^3(sec(x))^3 ?
The integral of (tan(x))^3(sec(x))^3 is (sec^5(x))/5-(sec^3(x))/3+C